Solve the following:

1) x^3 <= 16x
2) (5x/x^2+12x-8) >= 3

1.x^3 - 16x

= x(x^2 - 16)
= x(x^2 + 4x - 4x - 16)
= x[(x^2 + 4x) - (4x + 16)]
= x[x(x + 4) - 4(x + 4)]
= x(x + 4)(x - 4)

1. x^3 - 16x ≤ 0

x(x^2 - 16) ≤ 0
x(x-4)(x+4) ≤ 0

critical values are -4, 0, 4

making a quick sketch of y = x^3 - 16x
shows a standard cubic with x-intercepts of 0, ± 4
Where is the curve below the x-axis ?

x ≤ -4 or 0 ≤ x ≤ 4

2.
(5x/x^2+12x-8) >= 3
5x > 3x^2 + 36x - 24
3x^2 + 31x - 24 < 0

consider 3x^2 + 31x - 24
x = (-31 ± √1249)/6
= appr .7235 or -11.057

critical values: x = .7235 , x = -11.057

We also have to consider
x^2 + 12x - 8 , it cannot be zero
let's see for what values it is
solving, I got x = .63325 , x = -12.6332

looking at Wolfram's graph:
http://www.wolframalpha.com/input/?i=5x%2F(x%5E2%2B12x-8)+%3E%3D+3

we have
-12.6332 < x ≤ -11.057
OR
.63325 < x ≤ .7235

To solve the inequalities, we need to isolate the variable x. Here's how to solve each inequality step-by-step:

1) x^3 <= 16x

First, subtract 16x from both sides to bring all terms to one side:

x^3 - 16x <= 0

Now, we factor out x from the equation:

x(x^2 - 16) <= 0

Next, we factor the quadratic expression:

x(x + 4)(x - 4) <= 0

To determine when the expression is less than or equal to zero, we need to consider the sign of each term.

- If x < -4, then all three factors will be negative, which makes the expression positive.
- If -4 < x < 0, then the term (x + 4) becomes positive, while the other two terms remain negative, making the expression negative.
- If x = 0, the expression equals zero.
- If 0 < x < 4, the term (x - 4) becomes negative, while the other two terms remain positive, making the expression positive.
- If x > 4, then all three factors will be positive, which makes the expression positive.

We can summarize this information using a number line:

------------------o-----------------o-------------o----------

-4 0 4

Now, looking at the inequality, we see that it is asking for the values of x for which the expression is less than or equal to zero (x^3 ≤ 16x). Therefore, we are interested in the solution when the expression is negative (≤ 0).

From the number line, we see that the solution for x is:
x ≤ -4 or 0 ≤ x ≤ 4

2) (5x / (x^2 + 12x - 8)) ≥ 3

First, multiply both sides of the inequality by the denominator (x^2 + 12x - 8):

5x ≥ 3(x^2 + 12x - 8)

Expand the right side:

5x ≥ 3x^2 + 36x - 24

To solve this inequality, we need to bring all terms to one side and set it equal to zero:

3x^2 + 36x - 5x - 24 ≥ 0

3x^2 + 31x - 24 ≥ 0

Next, we can either factor the quadratic expression or use the quadratic formula to find the critical points. In this case, let's use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / (2a)

For the quadratic equation 3x^2 + 31x - 24 = 0, a = 3, b = 31, and c = -24.

Using the quadratic formula, we find two critical points:

x = (-31 ± √(31^2 - 4 * 3 * -24)) / (2 * 3)

Simplifying this expression gives us:

x = (-31 ± √(961 + 288)) / 6

x = (-31 ± √1249) / 6

Now, let's determine the sign of the equation for different ranges of x. We can make use of a number line to analyze this:

-------------------------------------------------------o---------o---------------

-10 -6 -1

To find the solution, we are interested in the solution when the expression is greater than or equal to zero (≥ 0).

From the number line, we see that the solution for x is:
x ≤ -10 or -1 ≤ x

Therefore, the solution to the inequality is x ≤ -10 or -1 ≤ x.